Question

Use completing the square to solve for x in the equation 2x^2-5x+1=3

Answer 1

x =
`(5)/(4)` ±
`(√(41))/(4)`

Explanation:

First, set the equation to equal zero.

`2x^2-5x+1=3`
-3 -3

`2x^2-5x-2=0`
Then, divide the equation by the coefficient of a (the value associated with
`x^2`) to equate a to 1.

`(2x^2-5x-2=0)/(2)`

`x^2-(5)/(2)x-1=0`
Next, identify the b in this equation. In quadratic form, it is the middle term with the x. Currently, the b equals
`-(5)/(2)`x. To create a perfect square trinomial, we must divide b by 2, then add the square of it to the equation.

`-((5)/(2))/(2)=-(5)/(4)`

`x^2-(5)/(2)x+(-(5)/(4))^2-1=0`
But you cannot simply add to an equation as that changes the fundamental truth of said equation. Therefore, we must also subtract
`(-(5)/(4))^2` from each side.

`(-(5)/(4))^2 = (25)/(16)`

`x^2-(5)/(2)x+(-(5)/(4))^2-1-(25)/(16) =0`
Now we can isolate our perfect square trinomial to the left side of the equation, then convert it to a squared binomial.

`x^2-(5)/(2)x+(-(5)/(4))^2-(41)/(16) =0`
+
`(41)/(16)` +
`(41)/(16)`

`x^2-(5)/(2)x+(-(5)/(4))^2 =(41)/(16)`

`(x-(5)/(4))^2 =(41)/(16)`
We can say we have properly completed the square now that we have our squared binomial. To solve this, we just have to isolate the x by taking the square root of each side and using the inverse operation of any constant on the x side.

`x-(5)/(4)` = ±
`\sqrt{(41)/(16)}`
+
`(5)/(4)` +
`(5)/(4)`
x =
`(5)/(4)` ±
`\sqrt{(41)/(16)}`
The square root of 16 is 4, so our final answer is...
x =
`(5)/(4)` ±
`(√(41))/(4)`
Which, if you're asking this due to Edge, is D.

Answer 2

Answer: did this help